Rutherford Scattering Cross-Section

Consider a particle of type $1$ , incident with relative velocity $u$ onto an ensemble of particles of type $2$ with number density $n_2$ . If $p_1({\mit\Omega})\,d{\mit\Omega}$ is the probability per unit time of the particle being scattered into the range of solid angle ${\mit\Omega}$ to ${\mit\Omega}+d{\mit\Omega}$ , then the differential scattering cross-section, $d\sigma/d{\mit\Omega}$ , is defined via (Reif 1965)

$$p_1({\mit\Omega})\,d{\mit\Omega} = n_2\,u\,\frac{d\sigma}{d{\mit\Omega}}\,d{\mit\Omega}.$$ (3.83)

Assuming that the scattering is azimuthally symmetric (i.e., symmetric in $\phi$ ), we can write $d{\mit\Omega}=2\pi\,\sin\chi\,d\chi$ . Now, the probability per unit time of a collision having an impact parameter in the range $b$ to $b+db$ is

$$p_1(b)\,db = n_2\,u\,2\pi\,b\,db.$$ (3.84)

Furthermore, we can write

$$p_1({\mit\Omega})\,\left\vert\frac{d{\mit\Omega}}{db}\right\vert = p_1(b),$$ (3.85)

provided that $\chi$ and $b$ are related according to the two-particle scattering law, Equation (3.82). It follows that

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{2\pi\,b}{\vert d{\mit\Omega}/db\vert}.$$ (3.86)

Equation (3.82) yields

$$\frac{d{\mit\Omega}}{db} = 2\pi\,\sin\chi\,\frac{d\chi}{db} = -2\... ...\left(\frac{4\pi\,\epsilon_0\,\mu_{12}\,u^2}{e_1\,e_2}\right)2\,\sin^2(\chi/2).$$ (3.87)

Finally, Equations (3.82), (3.86), and (3.87) can be combined to give the so-called Rutherford scattering cross-section,

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{1}{4}\left(\frac{e_1\,e_2}{4\pi\,\epsilon_0\,\mu_{12}\,u^2}\right)^2 \frac{1}{\sin^4(\chi/2)}.$$ (3.88)

It is immediately apparent, from the previous formula, that two-particle Coulomb collisions are dominated by small-angle (i.e., small $\chi$ ) scattering events.